> Gaetan Chenevier wrote: >> Is there an example of such a group? >> Yes, take the subfield of C generated by all the square roots of integers, >> call it K. > > OK, the final sentence of my posting on this was incorrect. Let me try to > understand the claim. If V is the direct sum of countably many copies of > the cyclic group of order 2, we can consider V as a vector space over the > field with two elements. Its dual vector space V* is the cartesian product > of countably many cyclic groups of order two. The double dual V** will be > bigger than V, which is the Pontryagin dual of V*. That's because V* is > uncountable, and, not very constructively, one can choose a basis and assign > values of a linear functional on it. Therefore there are many subgroups > of index 2 in V* that aren't closed in the product topology.
The Galois group you get by adding square roots of all the positive integers to Q is a vector space of dimension 2^countable over Z/2Z. You could find a basis for that, starting with the countable basis for the vector space which has all but finitely many components = . Call that basis B. Then you can fill that out to a basis B' for the whole vector space.
A basis for B is the automorphisms that send a square root of one prime to its negative, and leave everything else as it is. You could define a character on the Galois group to be 1 for elements of B, and 0 for all other elements of B'. The kernel of this character is a subgroup H of index 2.
Then suppose you have a subgroup of index 2 which is the fixed field of Q(sqrt(n)), some positive n. The character which has this subgroup as its kernel is 0 on some automorphism which sends the square root of some prime not a factor of n to its negative. So, H isn't such a subgroup.
I looked at the algebraic completion of Z/pZ, and for the Galois group of that over Z/pZ, subgroups of finite index do always correspond to fields.